CODEWITHSUNDEEP
python
Anmol Gulati
Anmol Gulati

Reputation: 83

Draw a grid like snake and ladder board game in python

I want to draw a normal terminal grid in python or you can just print the numbers in a snake and ladder way. The bottom line will be 1 to 10 in order, the line above that should 11 to 20 but in reverse order and continue this till 100. How can I achieve this using Python? A scalable solution to this would be much appreciated. Thanks.

EDIT : Ok, I came up with this solution. Is there any other way to achieve it. Maybe a more pythonic way or with less time and space complexity?

numbers = [i+1 for i in range(100)]
numbers = numbers[::-1]
levels = [i for i in numbers[::-10]]
is_reversed = False

for level in levels:

    if is_reversed:
        for number in reversed(numbers[level-1:level+9]):
            print('{:4}'.format(number), end='')
    else:
        for number in numbers[level-1:level+9]:
            print('{:4}'.format(number), end='')

    is_reversed = not is_reversed
    print()

Upvotes: 2

Views: 5078

Answers (3)

Manish V. Panchmatia
Manish V. Panchmatia

Reputation: 21

Here i is from 99 to 0 (both inclusive), the code is printing (i + 1) i.e. from 100 to 1

i // 10 will return identical value for 10 consecutive number. Based on weather this value is odd or even, the order of printing is decided.

Here i is from 99 to 0, i.e. descending order. So to print in reverse order, we need to jump with -10 for first column. for second column we need to jump with -9. However for second column the i is also reduced by 1. So we should jump with -8. Here -8 = -10 + 2 in generic way "jump = -10 + 2 (j th column)" Here column number j is from 0 to 9. It is remainder when we divide i with 10.

This is how, we get reverse order at alternate row

def snake_ladder():
    for i in range(99, -1, -1):
        if (i // 10) % 2 == 0:
            print("{0:4d}".format(i - 10 + 2 * (10 - (i % 10))), end=" ")
        else:
            print("{0:4d}".format(i + 1), end=" ")
        if i % 10 == 0:
            print("\r")

if __name__ == "__main__":
    snake_ladder()

output :

 100   99   98   97   96   95   94   93   92   91 
  81   82   83   84   85   86   87   88   89   90 
  80   79   78   77   76   75   74   73   72   71 
  61   62   63   64   65   66   67   68   69   70 
  60   59   58   57   56   55   54   53   52   51 
  41   42   43   44   45   46   47   48   49   50 
  40   39   38   37   36   35   34   33   32   31 
  21   22   23   24   25   26   27   28   29   30 
  20   19   18   17   16   15   14   13   12   11 
   1    2    3    4    5    6    7    8    9   10

Upvotes: 2

Eric Ed Lohmar
Eric Ed Lohmar

Reputation: 1922

The below uses a list comprehension to first make the list, then a second to reverse every other row. The key was the good ol' modulo %.

rows = [[f'{(n+1) + (i*10):4}' for n in range(10)] for i in range(10)]
rows = reversed([reversed(rows[i]) if i%2 else rows[i] for i in range(len(rows))])

for row in rows:
    print(' | '.join(row))

Result:

 100 |   99 |   98 |   97 |   96 |   95 |   94 |   93 |   92 |   91
  81 |   82 |   83 |   84 |   85 |   86 |   87 |   88 |   89 |   90
  80 |   79 |   78 |   77 |   76 |   75 |   74 |   73 |   72 |   71
  61 |   62 |   63 |   64 |   65 |   66 |   67 |   68 |   69 |   70
  60 |   59 |   58 |   57 |   56 |   55 |   54 |   53 |   52 |   51
  41 |   42 |   43 |   44 |   45 |   46 |   47 |   48 |   49 |   50
  40 |   39 |   38 |   37 |   36 |   35 |   34 |   33 |   32 |   31
  21 |   22 |   23 |   24 |   25 |   26 |   27 |   28 |   29 |   30
  20 |   19 |   18 |   17 |   16 |   15 |   14 |   13 |   12 |   11
   1 |    2 |    3 |    4 |    5 |    6 |    7 |    8 |    9 |   10

Edit:

A recent performance test showed me that enumerate is more performant than using range and indexing. That changes the second line to look like:

rows = reversed([reversed(row) if i%2 else row for i, row in enumerate(rows)])

Upvotes: 3

Joshua Nixon
Joshua Nixon

Reputation: 1427

for x in range(1, 101): # 1 - 100
    print ("%03d" % (101 - x,), end = " | ") # make all numbers 3 digits
    if x % 10 == 0: # every ten line output a new line
        print()

Output

100 | 099 | 098 | 097 | 096 | 095 | 094 | 093 | 092 | 091 | 
090 | 089 | 088 | 087 | 086 | 085 | 084 | 083 | 082 | 081 | 
080 | 079 | 078 | 077 | 076 | 075 | 074 | 073 | 072 | 071 | 
070 | 069 | 068 | 067 | 066 | 065 | 064 | 063 | 062 | 061 | 
060 | 059 | 058 | 057 | 056 | 055 | 054 | 053 | 052 | 051 | 
050 | 049 | 048 | 047 | 046 | 045 | 044 | 043 | 042 | 041 | 
040 | 039 | 038 | 037 | 036 | 035 | 034 | 033 | 032 | 031 | 
030 | 029 | 028 | 027 | 026 | 025 | 024 | 023 | 022 | 021 | 
020 | 019 | 018 | 017 | 016 | 015 | 014 | 013 | 012 | 011 | 
010 | 009 | 008 | 007 | 006 | 005 | 004 | 003 | 002 | 001 |

Upvotes: 1

Related Questions